leecode 解题总结：258. Add Digits

#include <iostream>
#include <stdio.h>
#include <vector>
#include <string>
using namespace std;
/*
问题:
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
For example:
Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.

Follow up:
Could you do it without any loop/recursion in O(1) runtime?

分析:这里的一位是指在十进制下面，只剩一位数
最简单的情况是：先分解然后相加每一位，
如果不使用循环或者递归，在O(1)时间完成。肯定只需要计算一遍
38,
3: 0011
8: 1000
异或操作:变为11
   1011

如果是33: 3 + 3 = 6,6+0=6这样不是永远
0011
0011
0000

输入:
33
38
1
100
输出:
6
2
1
1
*/

class Solution {
public:
	//将一个数的各个位进行拆分，返回拆分后的累加和。
	//拆分使用先 % 10 后 / 10的方法 
	int getResult(int num)
	{
		vector<int> result;
		int temp;
		do
		{
			temp = num % 10;
			result.push_back(temp);
			num /= 10;
		}while(num);
		//遍历求得累加和
		int sum = 0;
		if(result.empty())
		{
			return 0;
		}
		int size = result.size();
		for(int i = 0 ; i < size ; i++)
		{
			sum += result.at(i);
		}
		return sum;
	}

    int addDigits(int num) {
		int result = num;
		while(result >= 10)
		{
			result = getResult(result);
		}
		return result;
    }
};


void process()
{
	 int num;
	 Solution solution;
	 while(cin >> num )
	 {
		 int answer = solution.addDigits(num);
		 cout << answer << endl;
	 }
}

int main(int argc , char* argv[])
{
	process();
	getchar();
	return 0;
}